2d analysis of cantilver beam subjected to point load

8/19/2019 2D Analysis of Cantilver Beam Subjected to Point Load

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2D Analysis of Cantilever Beam Subjected to point Load

Introduction

This tutorial was created using ANSYS 7.0 to solve a simple 3D space frame problem.

Problem Description

The problem to be modeled in this eample is a simple !antilever beam Sub"ected to a point loadDimensions of the beam #ength $ %00mmDiameter $ &%mm'all Thic(ness $ &mm

Preprocessing: Defining te Problem

!" #ive te Simplified $ersion a %itle )such as *+erification ,odel*-.

tilit/ ,enu 1ile !hange Title

2" &nter 'eypoints

1or this simple eample2 these (e/points are the ends of the beam.

'e are going to define & (e/points for the simplified structure as given in the following table

( ey

poi

co

or

di

n

at

e

)

y

*


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1rom the *ANSYS ,ain ,enu* selectPreprocessor + ,odeling + Create + 'eypoints + In Active CS

-" .orm Lines

The two (e/points must now be connected to form a bar using a straight line.

Select Preprocessor + ,odeling+ Create + Lines + Lines + Straigt Line .

4ic( (e/point 56 )i.e. clic( on it-. t will now be mar(ed b/ a small /ellow bo.

Now pic( (e/point 5&. A permanent line will appear.

'hen /ou*re done2 clic( on *89* in the *!reate Straight #ine* window.

/" Define te %ype of &lement

t is now necessar/ to create elements on this line.

1rom the 4reprocessor ,enu2 select &lement %ype + Add0&dit0Delete.

!lic( on the *Add...* button. The following window will appear

1or this eample2 we will use the 3D elastic straight pipe element as selected in the above figure.Select the element shown and clic( *89*. You should see *T/pe 6 44:6;* in the *:lement T/pes*window.

!lic( on the *8ptions...* button in the *:lement T/pes* dialog bo. The following window will appear

0

0

0

2

%00

0

0


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!lic( and hold the 9; button )second from the bottom-2 and select *nclude 8utput* and clic( *89*.

This gives us etra force and moment output.

!lic( on *!lose* in the *:lement T/pes* dialog bo and close the *:lement T/pe* menu.

1" Define #eometric Properties

'e now need to specif/ geometric properties for our elements

n the 4reprocessor menu2 select eal Constants + Add0&dit0Delete

!lic( Add""" and select *T/pe 6 44:6;* )actuall/ it is alread/ selected-. !lic( on *89*.

:nter the following geometric properties

Outside diameterOD: 25

Wall thickness TKWALL: 2

This defines an outside pipe diameter of &%mm and a wall thic(ness of &mm.

!lic( on *89*.

*Set 6* now appears in the dialog bo. !lic( on *!lose* in the *


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!lose the *Define ,aterial ,odel =ehavior* 'indow.

'e are going to give the properties of Aluminum. :nter the following fieldEX

70000

PRX 0!""

Set these properties and clic( on *89*.

5" ,es Si*e n the 4reprocessor menu select ,esing + Si*e Cntrls + ,anualSi*e + Lines + All Lines

n the si>e *S?:* field2 enter the desired element length. 1or this eample we want an element length of &cm2 therefore2 enter *&0* )i.e &0mm- and then clic( *89*. Note that we have not /et meshed the geometr/2 we have simpl/ defined the element si>es.

)Alternativel/2 we could enter the number of divisions we want in the line. 1or an element lengthof &cm2 we would enter &% @ie &% divisions-.

67%&t is not necessar/ to mesh beam elements to obtain the correct solution. Bowever2 meshing is done inthis case so that we can obtain results )ie stress2 displacement- at intermediate positions on the beam.

8" ,es

Now the frame can be meshed. n the *4reprocessor* menu select ,esing + ,es + Lines and clic( *4ic( All* in the *,esh #ines*

'indow

9" Saving our ;or(


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This location is fied which means that all translational and rotational degrees of freedom )D81s-are constrained. Therefore2 select *All D81* b/ clic(ing on it and enter *0* in the +alue field and clic( *89*.

-" Apply Loads

As shown in the diagram2 there is a verticall/ downward load of 600N at the end of the bar

n the Structural menu2 select .orce0,oment + on 'eypoints.

Select the second 9e/point )right end of bar- and clic( *89* in the *Appl/ 1,* window.

!lic( on the *Direction of forcemom* at the top and select 1Y.

:nter a value of E600 in the *1orcemoment value* bo and clic( *89*.

The force will appear in the graphics window as a red arrow.

The applied loads and constraints should now appear as shown below.

/" Solving te System

'e now tell ANSYS to find the solution

Solution + Solve + Current LS


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Postprocessing: $ie=ing te esults

!" >and Calculations

Now2 since the purpose of this eercise was to verif/ the results E we need to calculate what we should find.

Deflection

The maimum deflection occurs at the end of the rod and was found to be ;.&mm as shown above.

Stress

The maimum stress occurs at the base of the rod and was found to be ;F.G,4a as shown above )pure bending stress-.

2" esults


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8bserve the value of the maimum deflection in the upper left hand corner )shown heresurrounded b/ a blue border for emphasis-. This is identical to that obtained via hand calculations.

Deflection

1or a more detailed version of the deflection of the beam2

1rom the *Ceneral 4ostproc* menu select Plot results + Contour Plot + 6odal Solution.

Select *D81 solution* and *S,*. #eave the other selections as the default values. !lic( *89*.


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You ma/ want to have a more useful scale2 which can be accomplished b/ going to the tilit/ ,enu and selecting Plot Controls + Style + Contours +


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Are these results what /ou epectedI Again2 the maimum deflection occurs at node &2 the right end of the rod. Also note that all the rotational and translational degrees of freedom were constrained to >ero at node 6.

f /ou wanted to save these results to a file2 use the mouse to go to the *1ile* menu )at the upper leftE hand corner of this list window- and select *Save as*.

Stresses

1or line elements )ie beams2 spars2 and pipes- /ou will need to use the &lement %able to gain access to derived data )ie stresses2 strains-.

1rom the #eneral Postprocessor menu select &lement %able + Define %able"""

!lic( on *Add...*


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As shown above2 in the *tem2!omp* boes in the above window2 select *Stress* and *von ,isesS:J+*

!lic( on *89* and close the *:lement Table Data* window.

4lot the Stresses b/ selecting Plot &lem %able in the :lement Table ,enu

The following window will appear. :nsure that *S:J+* is selected and clic( *89*

f /ou changed the contour intervals for the Displacement plot to Kser SpecifiedK /ou ma/ need to switch this bac( to KAuto calculatedK to obtain new values for +,N+,AL.


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Again2 select more appropriate intervals for the contour plot

#ist the Stresses 1rom the *:lement Table* menu2 select 4List &lem %able4 1rom the *#ist :lement Table Data* window which appears ensure *S:J+* is highlighted !lic( *89*

Note that a maimum stress of ;F.G6F ,4a occurs at the fied end of the beam as predicted anal/ticall/.

Bending Moment Diagrams

To further verif/ the simplified model2 a bending moment diagram can be created. 1irst2 let*s loo( at howANSYS defines each element. 4ipe 6; has & nodesM and 2 as shown in the following image.

To obtain the bending moment for this element2 the Element Table must be used. The :lement Table contains most of the data for the element including the bending moment data for each element at Node and Node . 1irst2 we need to obtain obtain the bending moment data.

#eneral Postproc + &lement %able + Define %able""" . !lic( *Add...*.


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n the window2A. :nter #$%ment as the *ser label for item* E this will give a name to the data=. Select *=/ seOuence num* in the tem bo!. Select *S,S!* in the first !omp bo

D. :nter &$#&'() in the second !omp bo:. !lic( *89*

This will save all of the bending moment data at the left hand side ) side- of each element. Now weneed to find the bending moment data at the right hand side ) side- of each element.

Again2 clic( *Add...* in the *:lement Table Data* window.A. :nter *$%ment as the *ser label for item* E again2 this will give a name to the data=. Same as above!. Same as aboveD. 1or step D2 enter &$#&'(+2 in the second !omp bo

:. !lic( *89*

!lic( *!lose* in the *:lement Table Data* window and close the *:lement Table* ,enu. Select Plot esults + Contour Plot + Line &lem es"""

1rom the *4lot #ineE:lement


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#ab2 and *,8,:NT* from the pull down menu for #ab. !lic( *89*. Note again that /ou can modif/ the intervals for the contour plot.

Now2 /ou can double chec( these solutions anal/ticall/. Note that the line between the and pointis a linear interpolation.

=efore the eplanation of the above steps2 enter hel, ,i,e+) in the command line as shown below and then hit enter.

=riefl/ read the ANSYS documentation which appears2 pa/ particular attention to the Tables near the end of the document )shown below-.

%able !" PIP&!3 Item? Se@uence 6umbers? and Definitions for te &%ABL& Commands

node

I

name

i

e

6

&

3

Def ini

,em

ber


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Note that S,S! ; )which we used to obtain the values at node - correspond to ,,8,? E the ,ember moment for node . The value of *e* varies with different :lement T/pes2 therefore /oumust chec( the ANSYS Documentation files for each element to determine the appropriate S,S!corresponding to the plot /ou wish to generate.

Command .ile ,ode of Solution

The above eample was solved using the Craphical ser nterface )or C- of ANSYS. This problem can also

been solved using the ANSYS command language interface. To see the benefits of the command line clear /our current file

1rom the

2d analysis of cantilver beam subjected to point load

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